9514 1404 393
(x, y) = (1, 1)
Step-by-step explanation:
About Algebra
All of Algebra is based on a few simple rules:
adding 0 doesn't change anythingmultiplying by 1 doesn't change anythingdoing the same thing to both sides of an equation doesn't change the values of the variablesanything can be substituted for its equal
When we solve equations by substitution, we make use of these facts.
__
About Substitution
Generally, substitution in equation solving is done for the purpose of eliminating a variable from the equation. That usually means you want to write an expression for one variable in terms of other variables.
Finding the Expression to Substitute
In this set of equations, the variables are x and y. We can substitute for y an expression involving x, or we can substitute for x an expression involving y. These equations are written in such a way that either method works as well as the other.
Let's choose to substitute for y. This means we need an equation of the form ...
y = <some expression involving x>
The first equation makes it easy to get that. If we add y (to both sides, see rule 3 above), we get ...
5x -y +y = 4 +y
5x = 4 + y . . . . . . . . . simplified
Subtracting 4 (from both sides) gets us the form we want.
5x -4 = 4 -4 + y
5x -4 = y . . . . . . by rule 4 above, we can use (5x-4) anyplace you see y
Using the Expression to Substitute
Now, we have an expression for y that we can use in the second equation. There, y is multiplied by 4, so we need to be sure to multiply the whole expression by 4:
-x + 4(5x -4) = 3 . . . . . . substituted for y in the second equation
-x +20x -16 = 3 . . . . . use the distributive property to eliminate parentheses
19x -16 = 3 . . . . . . collect terms
This has a constant (-16) where we don't want it, so we can add 16 (to both sides) to get rid of it.
19x -16 +16 = 3 +16
19x = 19 . . . . simplify
Now, we can divide by 19 (to both sides) to get x by itself.
(19x)/19 = 19/19
x = 1 . . . . almost done; we still need to find y
Finishing Up
Using the equation for y that we used for substitution, we can find y from our value of x.
y = 5x -4 = 5×1 -4 = 5-4
y = 1
So, the solution is x = 1, y = 1, or (x, y) = (1, 1).
Additional comment
In this set of equations, both variables (x and y) appear with coefficients of 1 (actually, -1). This makes it easy to solve for an expression that can be substituted for the variable. In systems of equations that you want to solve by substitution, it usually works well to start by looking for a variable with a coefficient of 1 (or -1). Use that to write the substitution expression as we did here.
__
Alternate Solution
If you use a variable with a coefficient other than 1 to write the substitution expression, you end up with fractions. That's OK if you're comfortable with arithmetic using fractions. It generally requires more care.
For example, we could solve the first equation for x:
x = (4 +y)/5
Substituting that into the second equation gives ...
-(4 +y)/5 +4y = 3
-4/5 -y/5 +4y = 3
(3 4/5)y = 3 4/5 . . . . . add 4/5 (to both sides), collect terms
Now, you get to divide by 3 4/5 to get ...
y = (3 4/5)/(3 4/5) = 1
Fortunately, in this case, we can see the numerator and denominator are identical, so the division result is 1. If it weren't you'd be faced with dividing mixed numbers--not always a student's first choice of fun math operations.
x = (4 +1)/5 = 1 . . . . finishing up
